## The Algebraic Hodge Theorem and the Fundamental Theorem of Elliptic Operators

** Statement of the Fundamental Theorem of Elliptic Operators**

**Definition. **Let and be inner product spaces. Let be a linear map. Then a linear map is called the formal adjoint of if for any and any .

**Lemma. ** (1) If a formal adjoint exists, it is unique. (2) If , then exists.

**Example.** The map defined by summing the coodinates has no formal adjoint, where is the colimit of .

If and are Hilbert spaces, then we have

**Theorem.** Any continuous linear map of Hilbert spaces has a formal adjoint.

**Example. ** Let be a differential operator. Suppose and have smooth inner product structures, then we have the “-inner products” on and , given by . Then has a formal adjoint . If we write locally as , then .

An elliptic operator is *self-adjoint* if .

We can now state the Fundamental Theorem of Elliptic Operators. Later in this entry we will give some corollaries and much later in the course we will outline a proof using the method of elliptic regularity.

**Fundamental Theorem of Elliptic Operators.** For a self-adjoint elliptic operator , there is an orthogonal decomposition with finite-dimensional.

It is important here that the manifold is closed.

** The algebraic Hodge theorem **

Suppose now we have (co)chain complex over or :

Give each an inner product. Assume each has a formal adjoint . Define the Laplacian . Then we have

**Lemma.** iff and .

*Proof. *Suppose . Then we have

and hence .

**Theorem. **Let be a (co)chain complex over a field . Then there exist decompositions such that the (co)chain complex can be written as

When or and is finite dimensional for each , setting and , the theorem above becomes a corollary of the following Algebraic Hodge Theorem:

**Algebraic Hodge Theorem.** Let be a (co)chain complex over or . Suppose that has inner product for each and that formal adjoint exists for each . Let , then

(1) TFAE: (a) , (b) and , (c) .

(2) .

(3) If is finite dimensional, then .

(4) If for any , then there are orthogonal decompositions

*Proof.* (1) (a) (b) (c) (a).

(2) Let , then .

(3) Show the inclusion in (2) is an equality by counting dimensions.

(4) It suffices to show the following orthogonal decomposition:

However easily we have

By checking the decomposition diagram above, we can obtain:

**Corollary.** is an isomorphism, where

**Corollary.** iff is an isomorphism for any .

** Wrapping up **

**Corollary. ** *Algebraic Wrapping up. For as above: * is an isomorphism. Hence is an isomorphism for all p.

This corollary “wraps up” a (co)chain complex into a single map.

Next, we consider wrapping up an ellliptic complex.

**Definition.** An e*lliptic complex of differential operators* is a cochain complex of differential operators

so that for all the associated symbol complex is exact.

If we define the symbol of differential operator of order by , then we have .

**Proposition.** Let be an elliptic complex of differential operators. Give and metrics for each . Then is an elliptic operator.

*Proof.* For any , since the complex is elliptic, we have the exact sequence

Thus, is an isomorphism.

Finally note that .

** Consequences of the fundamental theorem.**

We deduce the following corollaries of the Fundamental Theorem, the Algebraic Hodge Theorem, and Wrapping Up.

**Corollary.** Let be an elliptic complex of differential operators.

(1) For any , is finite dimensional.

(2) For any , .

(3) .

**Corollary.** If is an elliptic differential operator, then we have isomorphisms . Hence the kernel and the cokernel of an elliptic differential operator are finite dimensional.

**Corollary.** If is self-adjoint, then .

## Symbols

This post contains various definitions of the symbol of a differential operator. We will state a local version, then a global version and then we will finally view the symbol in its most abstract form: a section of a bundle over the total space of a cotangent bundle.

**Review of Local Definitions**

Let’s start by recalling that a differential operator of order on the manifold is is defined by:

where is smooth and if , then

Let and . The Symbol of , denoted by is then

A differential operator is said to be elliptic if for all and every we have that is invertible.

**Global definition of the Symbol**

Consider a globally defined differential operator

for and we want to define a linear map

in a coordinate free way.

With this in mind let and choose:

1. such that

2. such that

Then we define

Notice that even though this is a coordinate free definition of the symbol, it is still unclear how it changes in and . We will later see that is actually smooth on . Before this, we should prove that this definition is in fact independent on the choice of and .

* does not depend on *

**Claim 1** If is a smooth function such that and , then

*Proof:* For any differential operator , any section and any function ,

Setting we have

Induction on the order of and (3) will give us the result:

Let , then by definition

and so

Now assume the claim is true for every differential operator of order less than and suppose . By definition,

Thus, by induction

and notice that (3) gives us

so that

* does not depend on *

**Claim 2** Let be such that , then

*Proof:* It is easier if we use the easy direction of Peetre’s Theorem so that we can use the fact that is local, that is

equivalently

equivalently

So, since , we have as sought.

Let us finish the section with a short remark:

is homogeneous of degree in . That is, for every ,

Proof: Simply take instead of in the definition for .

**Local=Global**

**Lemma 1** For a differential operator of order , the two definitions of symbol coincide under the identification given by

*Proof:* Let . The function satisfies the conditions stated in the coordinate free definition of .
Let be the constant section , that is, for every .

Then (2) reads

where by (1)

Notice that here

since is a constant section.

Also notice that

1. for every :
This is because there is always a factor of in the expression for whenever .

2. :
This is a simple calculation.

Consolidating all the information we conclude

**Symbol as a section**

By consolidating definitions (*) and (1) of we get . Here is the bundle map and we are just looking at the diagram

To be explicit, if , then with . So

that is, and we are using the identification .

Smoothness follows from the smoothness of the local definition and the fact that both definitions coincide locally.

Finally, let

then we have

**Proposition 2** There is an exact sequence

Notice that this proposition (re)captures the fact that the symbol of an operator only `sees’ the `top’ degree of the operator.

**Fundamental Theorem of Elliptic Operators**

Now that we have a global definition of the symbol of a differential operator, we can state what it means for a differential operator to be elliptic. Namely, is elliptic if for every (i.e is in the complement of the zero section of the cotangent bundle), the map is invertible. The most important result involving elliptic operators is the following theorem:

**Theorem 3** Fundamental Theorem of Elliptic Operators

If is an elliptic differential operator over a compact manifold , then both and are finite dimensional vector spaces.

## Two beautiful theorems about C(X)

Let be a topological space. Let be the set of continuous functions from to . can also be thought of as set of smooth sections of the trivial bundle . Anyway, we get a contravariant functor

where is the category of . is the category of with and every sends to . The two beautiful to be discussed here are the following. *(Hewitt)*For , compact Hausdorff there is a bijection

*(Swan)*If is compact Hausdorff then taking sections gives a bijection from

These two beautiful *theorems*have some remarkable consequences If , are compact and Hausdorff then,

\textup{2} leads to the following result in –

is a consequence of the following Let be compact, Hausdorff topological space

- For ,
is a maximal ideal,

- If is a maximal ideal, then such that ,
- where is the set of all maximal ideals of a ring equipped with topology. The isomorphism takes to

*Proof:*

- Clearly is maximal as which is a field.
- Notice, if , then If is an such that for all in then for every , such that . Each there exists such that . Since is compact cover . Using which do not vanish on respectively, define
. Observe, . Define . Clearly and . Thus . Thus the only maximal ideals of is of the form for some .

- For any ideal of a ring define,
is the basis for all sets in the space under topology. The map

which sends

is already a bijection. All we need to show is

IF closed then define and \vspace{5pt}

IF be a basic closed set in , ie, for some then, define. Then is clearly a closed set and clearly .

*Proof:* *(of Theorem 1)* In fact the gives the map between the sets of the respective category.

**One-one**

Let . Then

, where . If then

by using bump functions near each point

**Onto**

Given a map , we induce a map

By \textup{5} we get a map

It is clear that . *Proof:* *( sketch of proof of theorem 2)*

Notice that Let be the map

G: isomorphism class of vector bundles over finitely generated C(X)-modules

where given a vector bundle

= smooth sections of .

Since is compact, any vector bundle is a of a trivial bundle of finite dimension, ie . Hence is a sub-module of due to the following isomorphism.

smooth sections on the trivial bundle

Thus is a finitely generated module. Moreover every bundle of finite dimension over a compact space has a complement, say , hence . Hence its projective. Given a finitely generated projective module over , say , find and a module , such that

Then define . This is a over . The proof is non-trivial and is a of .

## A local view of the global definition of a differential operator

Recall the global definition of a differential operator (of order *m*)

Also recall that differential operators are local, so that in particular, they induce (linear) operators between the spaces of sections over chart neighbourhoods in *X*. This allows us to reduce to the Euclidean case, and for the remainder of this discussion, we will assume (noting that the space of (smooth) sections of is naturally isomorphic to the space of (smooth) maps from , etc.)

**Proposition:** If *D* has the form of a “local differential operator” of order *m*, then

*Proof:*

Set as in the definition of a local differential operator. If then Leibniz’s differentiation rule yields are multi-indices, and sums of mulit-indices are taken term-wise, and .

We proceed by induction on *m*, the order of *D* (as a local D.O.). By the linearity of , it suffices to prove . Now, , which gives , establishing the base case. Now assume that all local differential operators of order at most are (global) differential operators of the same order. The Leibniz rule above gives:

Since —i.e., each in the sum has order at most , and so by the inductive hypothesis (and noticing that ), we find that QED.

Now we begin to prove the converse to the above proposition. For the purposes of the next lemma, we may relax the assumption that *D* is an operator between sections of vector bundles over Euclidean space.

**Lemma 1:** Let , then for every If every vanishes at a point .

*Proof:*

Again, by induction on *m*. If , and thus the base case is established. Now suppose the result for , and let . Now . By assumption, and so . On the other hand, by hypothesis, and so QED.

**Lemma 2: **Let Now, for every with the property that

The proof is left as an exercise. We give an outline of the inductive construction. To find , allow to act on the constant co-ordinate sections , and expand each in terms of the basis . Take to be the matrix thus determined—since *D* maps smooth vector fields to smooth vector fields, the co-ordinates of will be given by smooth functions of *x*. This gives a order approximation of *D*.

Now we take these sections and begin to multiply them by functions on *X* to obtain new sections—e.g., take where is the multi-index with only a single 1 in position *j*. In general, define the by for some multi-index . It is clear (or an exercise) that these matrices produce the desired result, and moreover, Lemma 1 shows that at most of these matrices will be non-zero (i.e., that the process terminates). Uniqueness is, of course, automatic.

**Theorem:** Differential operators are precisely local operators whose local form is that of a local differential operator.

*Proof:*

Because differential operators are local, we can immediately reduce to the Euclidean case. Showing “” was the content of the proposition, and so it suffices to show ““. Invoking the construction from lemma 2 (with *k = m*), let .

Consider the section . Recall Taylor’s Theorem: given , a polynomial of degree *m, *a neighbourhood , and for each multi-index such that for all . Choose take *m* to be the order of *D*, and apply Taylor’s Theorem to each individually to obtain on a neighbourhood . Take .

Now by locality, and so Hence, by construction, and so . On the other hand, , and so linearity reduces the question to showing .

Finally, are differential operators of order *m* (use the proposition to get this for ), and since smooth functions which all vanish at , we find by lemma 1, that , whence having been arbitrary. QED.

## A global definition of a differential operator

** Background **

and are smooth vector bundles over a smooth manifold . We have seen how to define differential operators locally, that is, over . How should we define *global* differential operators?

** Definitions **

Let us write . We define a Leibniz bracket by which acts on a section by . That is,

This is similar to the Lie bracket of two vector fields, .

We set and inductively define differential operators of order at most by requiring that provided for each .

Recall that for a section , its *support* is the closure of .

We would like to compare differential operators to local operators, those operators which only use local information in the following sense: An operator is local if, for each , .

Lemma 1A differential operator is local.

*Proof:* The only obvious way to prove this lemma is by induction. It is clearly true for : if is a section, then .

Now suppose that this is true for . Let , be arbitrarily chosen. Let be any open set for which . By Urysohn’s lemma, there is some smooth function with and . In particular, . (For brevity’s sake, call such an a “support function.”) Then since , we observe that

The support of the sum of two sections is the union of the supports of the sections since the sum is zero exactly when both sections are zero. Therefore,

The support of the product of two sections is the intersection of the supports of the sections, since the product is zero exactly when at least one of the sections is zero. Thus .

Additionally, by the inductive assumption. So

This containment holds for arbitrary containing . Therefore, the support of is contained in the intersection of the closures of these open sets:

One might naively wonder if this proves anything substantial: Are there ever any operators that aren’t local? In fact, yes; here’s an example. Let , , so that the sections of and both constitute smooth functions on . Let be a function with support on the northern hemisphere of , i.e., . Define an operator by

We see that cannot be local. Consider, say, a function with support only on the southern hemisphere of , so that . But by examining the integral , it’s clear that is not a subset of .

This gives intuition to the word “local:” a local operator acting on a section of determines based only on the behavior of in a neighborhood of , rather than on any global information. This corollary justifies the intuition: If on a neighborhood in two sections of agree, i.e., , then, restricted to , .

*Proof:* Put . Then and . Therefore, , so on we have that .

Since we’ve seen that differential operators are local, the natural next question is whether there are any non-differential local operators. This is answered by a theorem of Peetre, proved in the 1960s:

Theorem 2All local operators are differential operators.

The next question is whether in local coordinates differential operators can be represented in the form

The answer is “yes”:

Theorem 3All differential operators are, locally, differential operators.

## Smooth vector bundles and local coordinates

In this entry we remind the reader of the definition of a smooth vector bundle and give a local coordinates for a smooth section.

**Definition** [Milnor-Stasheff] A *rank n vector bundle* is a map and a vector space structure on for all so that and fiber-preserving homeo so that is a v.s. iso. To define a *smooth vector bundle* one requires that E and X are manifolds and the ‘s are diffeomorphisms.

Definition [Steenrod (see also Davis-Kirk)] A *rank n vector bundle* is a map and a collection of homeos so that

To define a *smooth vector bundle* one requires that X is a manifold and the are smooth.

A *smooth section* is a smooth map s.t. .

The vector space of smooth sections is a module over the ring of continuous functions .

e.g.

A smooth section of an -plane bundle over an -manifold is locally an element of . To make this precise and charts and . Then via

## Local differential operators

In this post we define a differential operator on Euclidean space and give some familiar examples.

For , the differential operator takes a smooth function on to a smooth function on . We call the **order** of this differential operator.

A linear map is a **(linear) differential operator of order ** if it is of the form , where is an matrix over . Given , the **symbol** is an matrix over such that , where .

We say that is **elliptic** if and for all , is an isomorphism of to itself.

**Examples**

Grad: ,

We have

Div: ,

We have

.

Curl: ,

We have

.

Exterior derivative ,

We have

Now, with respect to the ordered bases and for and , respectively, we may write this as

,

so by the previous example we see that .

Laplacian : ,

We have

.

All of the examples above are order 1 except for the Laplacian, which is order 2.

**Which of the above are elliptic?**

For dimensional reasons, the only candidates are , and . The first two are not elliptic since, for example, and has determinant . is elliptic, on the other hand, since .